Mathematics Question Thread

[RuiAce]

(Image removed from quote.)

[georgiia]

Thanks!!

[itssona]

hiii kinda stuck
how would the graph of y=1-cosx look?
domain x between -180 and 180

[RuiAce]

hiii kinda stuck
how would the graph of y=1-cosx look?
domain x between -180 and 180

[itssona]

ahh thank you Rui

[_____]

Stupid sexy hard inequality questions always throw me off.



Looking at v. Got P > 2r therefore r < P/2 but couldn't progress from there.

[mary123987]

Stupid sexy hard inequality questions always throw me off.

(Image removed from quote.)

Looking at v. Got P > 2r therefore r < P/2 but couldn't progress from there.

Hey in regards to this you must consider the fact that :
 θ  is ≤ 2π(as it is a circle)
now from part (i) we know that p=r( θ+2)
so p/r =2 +  θ
and p/r -2= θ
however we said that  θ<2π
so it can be wriiten as : p/r -2<2π
now just rearrange so that p/r <2(π+1)
p<2(π+1)r
p/2(π+1)<r
this is the left side of the inequality
Now for the left p=2r +rθ
since θ>0
p>2r
p/2>r
 now just combine to get
p/2(π+1)<r <p/2

Hope this helps

[_____]

Hey in regards to this you must consider the fact that :
 θ  is ≤ 2π(as it is a circle)
now from part (i) we know that p=r( θ+2)
so p/r =2 +  θ
and p/r -2= θ
however we said that  θ<2π
so it can be wriiten as : p/r -2<2π
now just rearrange so that p/r <2(π+1)
p<2(π+1)r
p/2(π+1)<r
this is the left side of the inequality
Now for the left p=2r +rθ
since θ>0
p>2r
p/2>r
 now just combine to get
p/2(π+1)<r <p/2

Hope this helps

Thanks!

[_____]

How do I find that T is (cosx, sinx) here? Apart from that I can complete the question.

[RuiAce]

How do I find that T is (cosx, sinx) here? Apart from that I can complete the question.

(Image removed from quote.)

This question was already addressed in the compilation.

[_____]

This question was already addressed in the compilation.

Didn't consider it a particularly hard one so I didn't look, mb.

Unfortunately that solution didn't help me much, as it says to use trig to find T which I what I'm struggling with. If I do this I get sinx = y/TO and cosx = x/TO (drawing a perpendicular line as suggested). How do I remove the denominator?

[RuiAce]

Didn't consider it a particularly hard one so I didn't look, mb.

Unfortunately that solution didn't help me much, as it says to use trig to find T which I what I'm struggling with. If I do this I get sinx = y/TO and cosx = x/TO (drawing a perpendicular line as suggested). How do I remove the denominator?

TO is the radius of the circle which has length 1

Also, I believe you meant sin(theta).

[_____]

TO is the radius of the circle which has length 1

Also, I believe you meant sin(theta).

Thanks. My own inability to see what's right in front of me is quite amazing 

[winstondarmawan]

Hello! Would appreciate help with the following:
https://scontent-syd2-1.xx.fbcdn.net/v/t34.0-12/21733200_1324623537666455_1924185311_n.png?oh=755e3ab03cb1b287d4ad471e3535d5c2&oe=59C00DE2
Apparently my friend got it from a 2U Trial paper. If this is not a 2U/3U question, then don't worry about it.
Thanks in advance.

[georgiia]

could I pls have help with both parts of the question? Thxx